Question

ssume kVp= 150kV. At this kVp, mass attenuation coefficients of bone and soft-tissue are0.14 cm2g−1and 0.15 cm2g−1, respectively. Assuming the densities of bone and soft-tissue are3.8 g cm−3and 1.5 g cm−3, respectively, calculate the numbers (relative to the incident number)of X-ray photons that will be transmitted through the left (bone and tissue) and right (tissueonly) halves of the body. (Points: 4)

Answer 1

We know that,

$I=I_oe^{-\mu R}$

With the dates that we have, we can calculate \mu for the bone and for the tissue, that is

$\mu_{tissue}=0.15*1.5cm^{-1}$

$\mu_{tissue}=0.225cm^{-1}$

And for the bone,

$\mu_{bone}=0.14*3.8cm^{-1}$

$\mu_{bone}=0.532cm^{-1}$

So, calculate I for left side,

For the tissue part,

$I_1 = I_0e^{-0.225*1}\\I_1=I_0e^{-0.225}$

FOr the bone part,

$I_2=I_0e^{-0.225}e^{-0.532*1}\\I_2=I_0e^{0.225+0.532}\\I_2=I_0e^{-0.757}$

And,

$I_3=I_0e^{-0.757}e^{-0.225}\\I_3=I_0e^{-0.982}$

Final Intensity for the left side is given by,

$I'=I_0e^{-0.982}\\\frac{I'}{I_0} = e^{-0.982}\\\frac{I'}{I_0}=0.374$

Moreover for the right side

$I'=I_0e^{-0.225*3}\\I'=I_0e^{-0.675}$

$\frac{I'}{I_0}=0.51$